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Math Help - Need help in squared differential EQn.

  1. #1
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    Need help in squared differential EQn.

    So my problem is this: "A non-negative function f has the property that its ordinate set over an arbitrary interval has an area proportional to the arc length of the graph above the interval. Find f"

    So i started with the integral of f and the definition of arc length in 2-space integral ( sqrt(1 + f ' (x)^2)) and then set them equal to each other. but then i get this:

    f(x) ^ 2 = 1 + f ' (x) ^ 2

    But I don't know how to solve this. Any help will be great.
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    So my problem is this: "A non-negative function f has the property that its ordinate set over an arbitrary interval has an area proportional to the arc length of the graph above the interval. Find f"

    So i started with the integral of f and the definition of arc length in 2-space integral ( sqrt(1 + f ' (x)^2)) and then set them equal to each other. but then i get this:

    f(x) ^ 2 = 1 + f ' (x) ^ 2

    But I don't know how to solve this. Any help will be great.
    You can do one of two things. First, solve for f' so

    f' = \pm \sqrt{f^2-1} which is now separable.

    Second, differentiate your ODE giving

    2 f f' = 2 f' f'' \; \text{or}\; f'(f''-f) = 0 which gives two cases f' = 0 or f''-f=0.

    Although I need to ask, where is your constant of proportionality?
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  3. #3
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    Thanks a lot for the help. And I thought the constants were suppose to some afterward in the general form of the solution.
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  4. #4
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    Quote Originally Posted by hashshashin715 View Post
    Thanks a lot for the help. And I thought the constants were suppose to some afterward in the general form of the solution.
    You do but the question talks about proportionality so you'll need a constant for that, i.e.

    f = k \sqrt{1 + f'^2}
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